2 Answers
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Let $r(t)$ be the radius of the patch at time $t$, measured in cm, with $t$ measure in seconds (why cm and seconds? Because the information we are given is in cm and in seconds, so there is no point in introducing unnecessary complications by changing units).

The area of the patch at time $t$ is then $A(t) = \pi \Bigl(r(t)\Bigr)^2$. Note that $A(t)$ is measured in $\text{cm}^2$.

We are told the rate of change of the Area over time; this is the derivative with respect to $t$; so we are told that
$$\frac{dA}{dt} = 5\frac{\text{cm}^2}{\text{s}}.$$
Why are the units $\text{cm}^2/\text{s}$? Because $\frac{dA}{dt}$ is measured in units of $A$ over units of $t$, in this case, square cm over seconds.

We are asked for: $r(20)$, and $r'(20)$.

How do we find them?

Well, we have an equation that relates $A$ and $r$. By taking derivatives, we obtain an equation that relates $A'$, $A$, $r'$ and $r$. We have:
$$\begin{align*}
A &= \pi r^2\\
\frac{d}{dt}(A) &= \frac{d}{dt}(\pi r^2)\\
\frac{dA}{dt} &= \pi\left( 2r\frac{dr}{dt}\right).
\end{align*}$$
Now, the area was $0$ when $t=0$, and grows at $5\ \text{cm}^2/\text{s}$, so after $20$ seconds the area will be $5(20) = 100$. Solving for $r(20)$ we get:
$$\begin{align*}
A(t) &= \pi(r(t))^2\\
A(20) &= \pi (r(20))^2\\
100 &=\pi (r(20))^2\\
\frac{100}{\pi} &= (r(20))^2\\
\sqrt{\frac{100}{\pi}} &= |r(20)|\\
\frac{10}{\sqrt{\pi}} &= r(20)\quad\text{(radius is positive)}\\
\frac{10\sqrt{\pi}}{\pi} &= r(20).
\end{align*}$$
So the radius at $t=20$ is $\frac{10\sqrt{\pi}}{\pi}$.

To find the rate of change of $r$ at $20$, we use the formula we found
$$\frac{dA}{dt} = \pi\left(2r\frac{dr}{dt}\right).$$
We know the value of $\frac{dA}{dt}$, and now we know the value of $r$ at $t=20$. So we just need to plug in and solve for $\frac{dr}{dt}\Bigm|_{t=20}$.